Question: A few families took a trip to an amusement park together. Tickets cost $$6.50$ each for adults and $$2.50$ each for kids, and the group paid $$41.00$ in total. There were $2$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Solution: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${6.5x+2.5y = 41}$ ${x = y-2}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-2}$ for $x$ in the first equation. ${6.5}{(y-2)}{+ 2.5y = 41}$ Simplify and solve for $y$ $ 6.5y-13 + 2.5y = 41 $ $ 9y-13 = 41 $ $ 9y = 54 $ $ y = \dfrac{54}{9} $ ${y = 6}$ Now that you know ${y = 6}$ , plug it back into ${x = y-2}$ to find $x$ ${x = }{(6)}{ - 2}$ ${x = 4}$ You can also plug ${y = 6}$ into ${6.5x+2.5y = 41}$ and get the same answer for $x$ ${6.5x + 2.5}{(6)}{= 41}$ ${x = 4}$ There were $4$ adults and $6$ kids.